Let $R,z > 0$ be positive real constants, and consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $$ f(v) = \frac{1}{\sqrt{(R+v)^2+z^2}}\ \mathrm{K}\!\left( \frac{4 R v}{(R+v)^2+z^2} \right)$$ where $\mathrm{K}(m)$ is the complete elliptic integral of the first kind, defined (using Wolfram Mathematica's convention) by $$ \mathrm{K}(m) = \int_0^{\pi/2} \frac{\mathrm{d} \theta}{\sqrt{1-m\sin^2 \theta}}. $$ Is $f$ an even function? Numerical tests suggest that it should be, but I can't find the right series of manipulations to demonstrate this analytically.

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    $\begingroup$ Okay, backtracking somewhat, one obvious thing to do is divide everything by $R^2$, and then you can relabel $v/R \mapsto v$, $z/R \mapsto z$ to get rid of $R$. Not much progress, but at least that leaves you fewer variables! $\endgroup$ – Chappers Apr 29 '15 at 1:00
  • $\begingroup$ Right, sorted it out (see below). $\endgroup$ – Chappers Apr 29 '15 at 1:24

Okay, cracked it. It's annoyingly simple in the end: $$ K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m\sin^2{\theta}}} \\ = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m(1-\cos^2{\theta})}} \\ = \int_0^{\pi/2} \frac{d\theta}{\sqrt{(1-m)+m\cos^2{\theta}}} \\ = \frac{1}{\sqrt{1-m}}\int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{m}{m-1}\cos^2{\theta}}} $$ and changing variables you conclude that $$ K(m) = \frac{1}{\sqrt{1-m}} K\left( \frac{m}{m-1} \right), $$ with caveats about square roots. Turns out you're fine, since $$ 1-m = 1- \frac{4v}{(1+v)^2+z^2} = \frac{(1+v)^2+z^2-4v}{(1+v)^2+z^2} = \frac{(1-v)^2+z^2}{(1+v)^2+z^2}, $$ clearly positive. And then you also have $$ \frac{m}{m-1} = -\frac{4v}{(1+v)^2+z^2} \frac{(1+v)^2+z^2}{(1-v)^2+z^2} = \frac{-4v}{(1-v)^2+z^2} $$

(Obviously I've set $R=1$ here, which is fine because I can just divide $v$ and $z$ by it.) Hence you end up with the evenness you asked for.

Remark: the identity I proved above is basically DLMF's 19.7.2's first equation, in different notation.

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  • $\begingroup$ That should be a $4Rv$ in the numerator of the argument of EllipticK, not $2Rv$. $\endgroup$ – David Zhang Apr 29 '15 at 0:17
  • $\begingroup$ Ah, that's more interesting... $\endgroup$ – Chappers Apr 29 '15 at 0:18
  • $\begingroup$ Fantastic! I think this is about as elegant a solution as I could have hoped for. $\endgroup$ – David Zhang Apr 29 '15 at 1:35
  • $\begingroup$ Yep, sometimes one just has to dig in and push the integrals around until they co-oporate :). Just don't ask me how to prove the AGM integral... $\endgroup$ – Chappers Apr 29 '15 at 1:39
  • $\begingroup$ For searching purposes: the relation you proved is called the imaginary modulus transformation. $\endgroup$ – J. M. isn't a mathematician Dec 30 '16 at 12:47

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